why i am getting sigsegv in hackerearth easy problem?

help
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1

Problem Link-https://www.hackerearth.com/practice/algorithms/graphs/graph-representation/practice-problems/algorithm/monk-in-the-real-estate/

Solution->I know i am accessing something wrong,Please suggest me changes.

#include
#include
#define li long int
using namespace std;
vector< vector > graph(10001);

int main()

{
li t=0;

cin>>t;

while(t!=0)

{

    li x=0;li y=0;

 li n;cin>>n;

li ans=0;

for(int i=1;i<=n;i++)

{

    cin>>x;

    cin>>y;
    
    if(graph[x][y]==0 )

   {

        graph[x].push_back(y);
        graph[y].push_back(x);

   }

}

for(int i=1;i<=n;i++)

{

       ans=ans+graph[i].size();

}

cout<<ans<<endl;

graph.clear();

t--;}

return 0;

}

1 Like
2

“if(graph[x][y]==0)”
You cannot access something that does not exist. g[x] exists but it is initially empty, hence you cannot access g[x][y].

Also please wrap your code in a <pre> block or provide a link to it, it is quite hard to read.

1 Like
3

but im getting error after entering 1 x and y(not after loop(i=1 to n)) why?

4

then i have to use array for already visited vertices?

5

There is no question of visiting vertices. You are simply taking the edges as input. You do not need the if statement there.

6

You are checking graph [x][y] even before taking any input in it. So it is not present there, that is why it is giving sigsegv

7

if input is 1,1 1,1 1,1 then why i need to insert 1,1 three times? basically im asking for how to check if this edges is already existed? please help me

8

To check if edge already exists you can use a map (or simply an array if limits allow) for this.

9

You will not usually need to do that because problems do not provide duplicate edges as input (what’s the point?). But if you are facing such a situation, use std::set instead of std::vector to store the neighbours of a vertex. A set does not allow duplicate elements, so even if you insert the same value multiple times, the set will store only one copy.

10

nice…i will

11

if i use list?